Pre algebra with variables
Introduction :
Pre algebra is the part of algebra which deals with solving basic algebra problems. The variables presents in algebraic expression is referred as pre algebra with variables. The variables present in the pre algebraic expression are represented with the help of alphabetic letters and the numbers are considered as constants. Pre algebra is mainly used to find the unknown variables value with the help of known values. The following are the example pre algebra problems with variables.
Pre algebra with variables example problems:
Example 1:
Solve the pre algebraic expression with variables.
5(-3m - 2) - (m - 3) = -4(4m + 5) + 10
Solution:
Given algebraic expression is
5(-3m - 2) - (m - 3) = -4(4m + 5) + 10
Multiplying the above factors
-15m - 10 - m + 3 = -16m - 20 +10
Now combining the above terms we get
-16m - 7 = -16m - 10
Add 16m + 7 on both sides of the above equation.
0m = 3
0 = 0
All real values are solution to the given equation.
Example 2:
Solve the pre algebraic expression with variables.
2(r -3) + 3s - 2(r -s -3) + 4
Solution:
Given algebraic expression
2(r -3) + 3s - 2(r -s -3) + 4
Multiplying the integer terms
= 2r - 6 + 3s -2r + 2s + 6 + 4
Now grouping the above terms we get
= 5s + 4 is the solution
Example 3:
Solve the pre algebraic expression with variables.
-4(e + 2) = e + 7
Solution:
Given expression is
-4(e + 2) = e + 7
Multiply factors in left term
-4e - 8 = e + 7
Add 8 on both sides
-4e - 8 + 8 = e + 7 + 8
Grouping the above terms
-4e = e + 15
Subtract e on both sides
-4e - e = e + 17 -e
Grouping the above terms
-5e = 15
Multiply -1 / 5 on both sides
e = -15/5
E = -3 is the solution.
Pre algebra with variables practice problems:
1) Solve the pre algebraic expression with variables.
-2(r - 1) – 4r - 1 = 3(5r + 2)
Answer: r = - 5/21 is the solution.
2) Solve the pre algebraic expression with variables.
-4(k + 2) = k + 9 + 2k
Answer: k = -17/7 is the solution.
Pre algebra is the part of algebra which deals with solving basic algebra problems. The variables presents in algebraic expression is referred as pre algebra with variables. The variables present in the pre algebraic expression are represented with the help of alphabetic letters and the numbers are considered as constants. Pre algebra is mainly used to find the unknown variables value with the help of known values. The following are the example pre algebra problems with variables.
Pre algebra with variables example problems:
Example 1:
Solve the pre algebraic expression with variables.
5(-3m - 2) - (m - 3) = -4(4m + 5) + 10
Solution:
Given algebraic expression is
5(-3m - 2) - (m - 3) = -4(4m + 5) + 10
Multiplying the above factors
-15m - 10 - m + 3 = -16m - 20 +10
Now combining the above terms we get
-16m - 7 = -16m - 10
Add 16m + 7 on both sides of the above equation.
0m = 3
0 = 0
All real values are solution to the given equation.
Example 2:
Solve the pre algebraic expression with variables.
2(r -3) + 3s - 2(r -s -3) + 4
Solution:
Given algebraic expression
2(r -3) + 3s - 2(r -s -3) + 4
Multiplying the integer terms
= 2r - 6 + 3s -2r + 2s + 6 + 4
Now grouping the above terms we get
= 5s + 4 is the solution
Example 3:
Solve the pre algebraic expression with variables.
-4(e + 2) = e + 7
Solution:
Given expression is
-4(e + 2) = e + 7
Multiply factors in left term
-4e - 8 = e + 7
Add 8 on both sides
-4e - 8 + 8 = e + 7 + 8
Grouping the above terms
-4e = e + 15
Subtract e on both sides
-4e - e = e + 17 -e
Grouping the above terms
-5e = 15
Multiply -1 / 5 on both sides
e = -15/5
E = -3 is the solution.
Pre algebra with variables practice problems:
1) Solve the pre algebraic expression with variables.
-2(r - 1) – 4r - 1 = 3(5r + 2)
Answer: r = - 5/21 is the solution.
2) Solve the pre algebraic expression with variables.
-4(k + 2) = k + 9 + 2k
Answer: k = -17/7 is the solution.